$$\begin{array}{l}{R_{1}=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}} \\ {R_{2}=\{(1,1),(2,2)\}} \\ {R_{3}=\{(1,2),(2,3),(3,1)\}}\end{array}$$
$$\begin{array}{l}{R_{4}=\{(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)\}} \\ {R_{5}=\{(1,1),(2,2),(3,3),(1,2),(2,3),(3,1)\}} \\ {R_{6}=\{(1,2),(2,1)\}}\end{array}$$
My answer
In order for $R_i$ to be an equivalence relation it should meet four criteria
- $R_i$ should be a subset of $\{1,2,3\} \times \{1,2,3\}$
- $R_i$ should be reflexive. symmetric and transitive.
My conclusion is that none of the relations above are an equivalence relation. But the answer key is $R_1$. The problem is that $3$ is not related to any of the others and one cannot see that $R_1$ is transitive.
Q1: Why am I wrong?
Q2: Why isn't $R_5$ an equivalence relation?
Best Answer
For transitivity you need for all $a,b,c\in \{1,2,3\}$: If $a$ is related to $b$ and $b$ is related to $c$, then $a$ is related to $c$. You can indeed check for $R_{1}$ that this is true, since if you pick $a = 3$, then the statement is trivial. Since it only satisfies the "if" condition if you pick $b = 3$, since $3$ is only related to $3$.
Clarification: The thing you have to check for transitivity is for every $(a,b) $ and $(b,c)$ in $R_{1}$ you also have $(a,c)$ in $R_{1}$. Maybe this way of phrasing transitivity helps you to understand it better.